Given a covariant riemannian metric of certain sobolev class (i.e. with square-integrable weak derivatives up to a large enough integer order) on a (compact, if necessary) finite-dimensional smooth manifold, I would like to know the answers to the following related questions:

(1) does its inverse (i.e. the contra-variant type) belong to the same sobolev class?

(2) If a sequence of (covariant) metrics converges in the given sobolev norm to a limit metric, does it follow that the sequence of corresponding inverse metrics converges in the same sobolev norm to the inverse of the above limit metric?

***One of the things that I am iffy about: the inverse of a non-degenerate matrix involves the reciprocal of its determinant. I can believe that the determinant of a metric is a nowhere vanishing sobolev function but I am not sure whether it holds true for its reciprocal.

$\begingroup$If you have large enough order to get, by Sobolev embedding, that the metric is in $L^\infty$, then the question is an issue of arithmetic. // In general, your question is basically the same as asking "if the function $f\in W^{s,p}$ then is the function $1/f \in W^{s,p}$?"$\endgroup$
– Willie WongAug 14 '18 at 20:05

$\begingroup$These are mostly discussed in Section 5.6 of L.C. Evans' Partial Differential Equations; the statement that $W^{k,p}$ is an algebra is relegated to an exercise at the end of the chapter.$\endgroup$
– Willie WongAug 14 '18 at 20:31

$\begingroup$For some reason, I find it easier to understand the general case, rather than for just the function $t \mapsto 1/t$. Instead, assume that you have a smooth function $\Phi: (a,b) \rightarrow \mathbb{R}$ and you want to prove that if $f$ lies in a Sobolev space and its image lies in $(a,b)$, then $\Phi\circ f$ lies in the same Sobolev space.$\endgroup$
– Deane YangAug 14 '18 at 22:56